Question

Let f(x)=(x^2+3x-4) and g(x)=(x+4)

a. FInd f(times)g and state the domain
b.Find f/g and state the domain

Answer 1

a. would be set up as
(x^2+3x-4)(x+4) right?
and b. would be
(x^2+3x-4)/(x+4)

Answer 2

yes

Answer 3

what do i do from there?

Answer 4

you find the domain - x-coordinates. Just plug in a number for x several times and you have your answer

Answer 5

No. For A, we have f * g, for which all values are defined. The domain is the set of all real numbers. However, when we have \(\frac{x^2 + 3x - 4}{x+4}\), there is a point where it is undefined. It will be undefined you have division by zero; when is the denominator, \(x+4\), equal to 0?

Answer 6

I'm not sure what your asking @bloopman
for a, i'm just solving for x correct?

Answer 7

im just confused with b mainly

Answer 8

For a, the answer is all real numbers. There is no place where f(x) will not be defined. But how in god's name could you have \(\frac{x^2 + 3x -4}{0}\)? There is a place where \(x + 4 = 0\). Where is that place?

Answer 9

I have no idea...?

Answer 10

okay so for a, do I just distribute whats inside the first parenthesis to the second?

Answer 11

No. Why would you need to? Do you know what the domain of a function is?

Answer 12

Also, how do you solve \(x + 4 = 0\)? Functional analysis shouldn't be in your question if you're not sure how to solve that.

Answer 13

I have no idea how to solve it, which is why I'm asking,
I've looked back in my notes, and this type of question is nowhere but it's in the review quiz so I'm pretty sure it's important for the exam coming up..